Question

The oxidation numbers of phosphorus in $Ba(H_{2}P{O}_2{)}_2$ and xenon in $N{a}_{4}Xe{O}_6$ respectively are:

• A. $+3$ and $+4$
• B. $+2$ and $+6$
• C. $+1$ and $+8$
• D. $-1$ and $-6$

Answer 1

The correct option is C $+1$ and $+8$

$B{a}^{2+}\stackrel{+1}{(H_2}\stackrel{x-2}{P{O}_2}{)}_2$
$\therefore 2+2[2\times (+1)+x+2\times (-2\left)\right]=0$
or $2+4+2x-8=0$ or $x=+1$
and $\stackrel{+1}{N{a}_4}\stackrel{x-2}{Xe{O}_6}$
$\therefore 4\times 1+x+6\times (-2)=0$ or $x=+8$.